2023 AMC 8 Problem 18

Below is the video solution and professionally curated solution for Problem 18 of the 2023 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 8 solutions, or check the answer key.

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Concepts:Diophantine Equationmodular arithmeticoptimization

Difficulty rating: 1740

18.

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump 55 pads to the right or 33 pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located 20232023 pads to the right of her starting position?

405405

407407

409409

411411

413413

Video solution:
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Written solution:

Let rr be the number of right jumps and ll be the number of left jumps. We need 5r3l=2023. 5r-3l=2023. Modulo 55, this gives 3l3(mod5)-3l\equiv3\pmod5, so l4(mod5)l\equiv4\pmod5.

To minimize the total number of jumps, use the smallest possible ll, namely l=4l=4. Then 5r12=20235r-12=2023, so r=407r=407.

The fewest number of jumps is 407+4=411407+4=411.

Thus, D is the correct answer.

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